polynomials - How to set/adjust Precision for an iterative calculation?

How should I restructure this code? I generate a high-order polynomial poly with integer coefficients, then find roots and divide them out of poly one at a time. I want FindRoot[] to use a very high precision on the first pass but then just use the precision of the inputs afterward.

z[n_, c_] := If[n > 0, z[n - 1, c]^2 + c, c];
poly = PolynomialQuotient[z[10, c] - z[6, c], 1 + c^2, c];
rts = {};
Do[Print[Precision[poly]];
 aa = FindRoot[poly, {c, I}, WorkingPrecision -> Min[500, Floor[Precision[poly]]]];
 AppendTo[rts, c /. aa];
 poly = PolynomialQuotient[poly, ((z - c)*(z - Conjugate[c]) /. aa) /. z -> c, c], {j, 1, 5}]

(* [Infinity] 319.649 135.906 0.00598703 *)

I'm losing so much precision on each pass that I can only get a few roots. I'm trying to get the roots closest to i without having to find all the roots. I encountered this precision problem while trying to fix this other problem. I'm also not clear on WHY the polynomial division loses precision so quickly.

Answer

Precision is preserved, and error messages are eliminated by replacing the second instance of PolynomialQuotient by a simple divide.

poly = poly/ (((z - c)*(z - Conjugate[c]) /. aa) /. z -> c)

The only apparent difference is that PolynomialQuotient discards any remainder, and there is a remainder unless c is exact. In some way that I do not understand, discarding the remainder must reduce the precision of poly. So, this may not be a particularly satisfying answer, but it does produce accurate roots for a WorkingPrecision as low as Min[190, Floor[Precision[poly]]]. Replace 190 by 187, however, and the precision of poly gradually decreases to 185, whereupon the Jacobian becomes singular. For completeness, rts[[5]] for 190 is

(* -0.01660571703737496762392836921351877966202681662442570475568531876233767935059224313655985208958046768635767201058325661435694155300994327878649425388424142524075655339495830479365786600752927
 + 1.006001836522824948881217257805018657146542248017733702434346228688166066459126472333560301933200993465308529209500537579261396301606919993043401875320490162101139793328283528897296023734789 I *)

Comments

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1....

Blog

Search This Blog